Constructed metric problems train the habit of checking the axioms carefully. Some metrics are built by adding a fixed penalty when points lie in different parts of a set. Others are pullbacks of known metrics through injective maps. The common theme is that identity of indiscernibles and the triangle inequality must be protected. :::solved-problem Let $X=\mathbb{R}$ and define $$\begin{equation*} d(x,y)= \begin{cases} |x-y|+1, & \text{exactly one of }x,y\text{ is strictly positive},\\ |x-y|, & \text{otherwise}. \end{cases} \end{equation*}$$ Prove that $d$ is a metric. ::: :::solution Let $d$ be defined as stated. Non-negativity and symmetry follow from the definition. If $d(x,y)=0$, then $|x-y|=0$, so $x=y$. Conversely, $x=y$ gives $d(x,y)=0$. For the triangle inequality, if $x$ and $y$ are separated by strict positivity, then at least one of the pairs $(x,z)$ and $(z,y)$ is also separated by strict positivity. Thus the added $1$ appears on the right side, and the usual triangle inequality gives the result. Hence $d$ is a metric. $\square$ ::: :::solved-problem Let $(X,d)$ be a metric space and let $f:X\to X$ be one-to-one. Define $D(x,y)=d(f(x),f(y))$. Prove that $D$ is a metric on $X$. ::: :::solution Let $D(x,y)=d(f(x),f(y))$. The first axiom follows from non-negativity of $d$. If $D(x,y)=0$, then $f(x)=f(y)$, and injectivity gives $x=y$. Symmetry follows from symmetry of $d$. For the triangle inequality, $$\begin{equation*} D(x,y)=d(f(x),f(y))\leq d(f(x),f(z))+d(f(z),f(y))=D(x,z)+D(z,y). \end{equation*}$$ Hence $D$ is a metric. $\square$ ::: :::solved-problem Let $(X,d)$ be a metric space and let $X=U\cup V$, where $U$ and $V$ are disjoint nonempty subsets. Define $D(x,y)=d(x,y)+1$ if exactly one of $x,y$ belongs to $U$, and $D(x,y)=d(x,y)$ otherwise. Prove that $D$ is a metric. ::: :::solution Let $D$ be defined as stated. The first three axioms follow from the definition and the corresponding axioms for $d$. For the triangle inequality, if $x$ and $y$ lie in the same part, then $D(x,y)=d(x,y)\leq D(x,z)+D(z,y)$. If $x$ and $y$ lie in different parts, then at least one of $(x,z)$ and $(z,y)$ crosses the partition, so the added $1$ appears on the right side. The triangle inequality for $d$ then gives $D(x,y)\leq D(x,z)+D(z,y)$. Hence $D$ is a metric. $\square$ ::: :::solved-problem Let $X=\mathbb{R}^2$. For $x=(x_1,x_2)$ and $y=(y_1,y_2)$, define $$\begin{equation*} d(x,y)= \begin{cases} |x_1-y_1|, & x_2=y_2,\\ |x_1|+|x_2-y_2|+|y_1|, & x_2\neq y_2. \end{cases} \end{equation*}$$ Show that $d$ is a metric. ::: :::solution Let $d$ be defined as stated. The first three axioms are direct. The formula measures movement along a horizontal line, or movement from the first point to the vertical axis, then vertically, and then horizontally to the second point. Any path through an intermediate point cannot be shorter than the direct expression. Therefore the triangle inequality holds. Hence $d$ is a metric on $\mathbb{R}^2$. $\square$ ::: :::solved-problem Let $X=\mathbb{R}$ and define $$\begin{equation*} d(x,y)= \begin{cases} 0, & x=y,\\ |x|+|x-y|+|y|, & x\neq y. \end{cases} \end{equation*}$$ Show that $d$ is a metric. ::: :::solution Let $d$ be defined as stated. Non-negativity, symmetry, and identity of indiscernibles follow directly. If $x\neq y$, then the usual triangle inequality gives $|x-y|\leq |x-z|+|z-y|$. The remaining terms in $d(x,z)+d(z,y)$ are non-negative and include the necessary endpoint contributions, with equality in cases where $z=x$ or $z=y$. Hence $d(x,y)\leq d(x,z)+d(z,y)$. Thus $d$ is a metric. $\square$ ::: :::exercise [1] Why must the pullback map be one-to-one? [2] In the partition construction, why does one shorter pair cross the partition when the endpoints are in different parts? [3] Verify identity of indiscernibles for the modified real-line metric. ::: :::answer [1] Without injectivity, distinct points could have the same image and zero pullback distance. [2] The intermediate point lies in exactly one part. [3] The distance is zero by definition when $x=y$ and is positive when $x\neq y$. ::: :::faq[Frequently Asked Questions] Q: Why do many constructed metrics add $1$? A: The added term creates a fixed separation between different regions. Q: Why is injectivity needed in pullback metrics? A: It preserves identity of indiscernibles. Q: What is the main axiom to test in constructed metrics? A: The triangle inequality. ::: :::call-to-action[Revise the Metric Space Foundations] subtitle: Review metric axioms, standard examples, convergence, Cauchy sequences, completeness, and completion before moving to topology. button: Section Overview | /functional-analysis/metric-spaces/introduction-to-metric-spaces button-ghost: First Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/metric-axioms-and-subspace-metrics :::